High-quality black-and-white photographs of large old machines and tools

[mklotz]

The line about hat tipping reminded me of a Scientific American article, early seventies?, about a hat-tipping device patent of years past. Gentlemen could perform a hat-tip by squeezing a rubber bulb in a pants pocket. Airline was routed under back of jacket, along neck, behind ear and into hat. Ridiculous in the extreme, but I think it was granted, year unknown. Drawings showed bowler hats.

I remember that article. There are a lot of really weird patents out there; thankfully, most of them never made it to production.

[Jon]

One example: https://pdfpiw.uspto.gov/.piw?PageNum=0&docid=00556248

[Jon]

Captioned as: "Steam tractor burning some oil."

Fullsize image: https://diqn32j8nouaz.cloudfront.net...l_fullsize.jpg

[Jon]

Robertson tube straightener. I believe this machine is from around the 1940s.

Fullsize image: https://diqn32j8nouaz.cloudfront.net...r_fullsize.jpg

[Frank S]

Robertson tube straightener. I believe this machine is from around the 1940s.

Fullsize image: https://diqn32j8nouaz.cloudfront.net...r_fullsize.jpg

My first thought when I saw the guy standing there was he looked like Liam Neeson standing there the resemblance is striking

[Toolmaker51]

As if I need another project; salvaged a decent amount of wiring from my building. It's been coiled but should be on spools. They were only 110v circuits and look good as new, so will reuse them as such. Goal of respooling is I have a measuring wheel, to avoid pulling what turns up too short.
I have plenty conveyor skate wheels, a length of good sized angle iron, just need to cobble up axles and adjusting screws.
Any suggestions?

[Big Sexy]

As if I need another project; salvaged a decent amount of wiring from my building. It's been coiled but should be on spools. They were only 110v circuits and look good as new, so will reuse them as such. Goal of respooling is I have a measuring wheel, to avoid pulling what turns up too short.
I have plenty conveyor skate wheels, a length of good sized angle iron, just need to cobble up axles and adjusting screws.
Any suggestions?

Well there a couple of different ways you can do it with less work. First and easiest way is to count the number of loops and multiple by the diameter of the coils. The second is more accurate but does require a decent quality multimeter. You take a resistance measurement of the coil and then determine what size of wire you have. Then look up the known resistance for the size of wire you have. Once you know the resistance per length of wire you then multiply or divide your resistance by the known resistance for length. I think that’s how the second is done. It’s been a while and would need numbers in front of me to refresh my memory lol. But it is still easily done.

[mklotz]

Well there a couple of different ways you can do it with less work. First and easiest way is to count the number of loops and multiple by the diameter of the coils. The second is more accurate but does require a decent quality multimeter. You take a resistance measurement of the coil and then determine what size of wire you have. Then look up the known resistance for the size of wire you have. Once you know the resistance per length of wire you then multiply or divide your resistance by the known resistance for length. I think that’s how the second is done. It’s been a while and would need numbers in front of me to refresh my memory lol. But it is still easily done.

Rather than looking up the resistance/length value for the wire, you'll get better accuracy by measuring resistance of a known length of the subject wire. You'll need a very accurate ohm meter; most are not terribly accurate for measuring small resistances. Very small resistances are normally measured with a current bridge arrangement...

https://en.wikipedia.org/wiki/Wheatstone_bridge

Let:

r = resistance of length l of wire
R = resistance of coil of wire
L = length of coil of wire

then:

r / l = R / L ; assume resistance per unit length is constant over length of wire

or:

L = R / (r / l) = (R / r) * l

Taking the derivative of 'L' wrt to 'r'...

dL = - (R / r^2) * l * dr = - (L / r) * dr

which will allow you to estimate the error in 'L', dL, for a given error in the measurement of 'r', dr.

[Toolmaker51]

I like the multi-meter solution. It will also start screening broken conductors.
1st priority remains building the straightener to respool the wire for storage an eventual use.

[Frank S]

I like the multi-meter solution. It will also start screening broken conductors.
1st priority remains building the straightener to respool the wire for storage an eventual use.

Also with the multimeter if you know the length & gage +the ohms the wire should be, if they are off by a large margin it means the wire has been overloaded not fit for installing in a circuit